On Sun, Feb 21, 2021 at 01:43:03PM -0500, Jeff Layton wrote:
> On Sun, 2021-02-21 at 16:52 +0000, Al Viro wrote:
> > On Sat, Feb 20, 2021 at 01:32:50AM -0500, Luo Longjun wrote:
> > > + list_for_each_entry(bfl, &fl->fl_blocked_requests, fl_blocked_member)
> > > +         __locks_show(f, bfl, level + 1);
> > 
> > Er...  What's the maximal depth, again?  Kernel stack is very much finite...
> 
> Ooof, good point. I don't think there is a maximal depth on the tree
> itself. If you do want to do something like this, then you'd need to
> impose a hard limit on the recursion somehow.

I think all you need to do is something like: follow the first entry of
fl_blocked_requests, printing as you go, until you get down to lock with
empty fl_blocked_requests (a leaf of the tree).  When you get to a leaf,
print, then follow fl_blocker back up and look for your parent's next
sibling on its fl_blocked_requests list.  If there are no more siblings,
continue up to your grandparent, etc.

It's the traverse-a-maze-by-always-turning-left algorithm applied to a
tree.  I think we do it elsewhere in the VFS.

You also need an integer that keeps track of your current indent depth.
But you don't need a stack.

?

--b.

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